some elementary linear algebra

Today I encounter sometihing strange, at least to me in elementary linear algebra.

Suppose that there are two real symmetric matrices, A,B and A is definite positive. Then there is a square matrix C such that C'C=A,C'bC=B where C' is the transpose of C.

At first glance, I think of diagonilization in linear algebra. And so this result strike me: we can always diagonalize two symmetric matrices simultaneuously!

Then searching on the Internet for a while, I found that I was wrong. This is not diagonalization. This is related more to (symmetric) bilinear forms than to linear transformations. This is a very important difference.

Having found the starting point, things become easier to understand. We can give \mathbb{R}^n with the standard inner product <,> and define f_A(x,y)=<x,y>_A=<Ax,y>. This is a symmetric bilinear form, and is non-degenerate since A is not. So there is an orthogonal basis for f_A, that is to say, under this basis e_1,...,e_n, f_A looks like the standard inner product, f_A(\sum_ix_ie_i,\sum_jy_je_j)=\sum_ix_iy_i. And thus with the matrix C=(e_1 e_2 ... e_n), we have that A=C'C. We can thus rewrite the bilinear form as F_A((x_i),(y_i))=f_A(\sum_ix_ie_i,\sum_jy_jej)=<(x_i),(y_i)>. So, under this expression, we can define another bilnear form, F_B((x_i),(y_i))=<B(x_i),(y_i)>. So, since F_B is symmeytic, we can find an orthonormal basis, E_1,...,E_n such that under this new basis, we have F_B(\sum_iX_iE_i,\sum_jY_jE_j)=\sum_ib_iX_iY_i. We denote the orthogonal matrix as E=(E_1 ... E_n).

All having been said, let’s look at what happens here. in the basis e_i, f_A is the usual inner product. So we have that A=C'C. Under the basis, E_i, F_B becomes a diagonal matrix D=diag(b_i), so we have that B=(EC)'D(EC). Note that we also have A=(EC)'(EC). So at last we find a matrix M=EC such that A=M'M and B=M'DM.

One thing to keep in mind if there is something worth to do so: bilinear form and linear transformation are different.

Note that the result nevers says that two symmetric matrices can be simultaneuouly diagonalized even if there is at least one of them is definite positive. This will mean that they commute with each other. But it is not the case in general. we can found some very simple counterexamples.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s